jcst.net
当前位置:首页 >> 设数列{An}(n=1,2,…)是等差数列,且公差为D,... >>

设数列{An}(n=1,2,…)是等差数列,且公差为D,...

(1)证明:an=4+(n-1)?2=2n+2,对任意的m,n∈N*,有am+an=(2m+2)+(2n+2)=2(m+n+1)+2,∵m+n+1∈N+,于是,令p=m+n+1,则有ap=2p+2∈{an}.∴该数列是“封闭数列”.(2)∵a1=-5,a2=-3,∴a1+a2=-8,令an=a1+a2=-8,∴2n-7=-8,n=-12?N+,所以...

(1)∵an+1=2Sn+1,∴当n≥2时,an=2Sn-1+1两式相减得:an+1=3an(n≥2)又a2=2a1+1=3=3a1,∴an+1=3an(n∈N*).∴数列{an}是以1为首项,3为公比的等比数列,∴an=3n-1.又b1=b2-d=5-2=3,∴bn=b1+(n-1)d=2n-1.(2)an?bn=(2n+1)?3n?1令Tn=3×1+5×...

解: am=Sm-S(m-1)=0-(-9)=9 Sm=(a1+am)m/2=0 a1+am=0 a1=-am=-9 又am=a1+(m-1)d d=(am-a1)/(m-1)=[9-(-9)]/(m-1)=18/(m-1) m>3,18/(m-1)

由数列{an}是公差为d,且首项为a0=d的等差数列得:an=a0+(n+1-1)d=(n+1)d;∴Sn+1=a0C0n+a1C1n+…+anCnn,又Sn+1=anCnn+an?1Cn?1n+…+a0C0n=anC0n+an?1C1n+…+a0Cnn,∴2Sn+1=(a0+an)C 0n+(a1+an?1)C1n+…+(an+a0)Cnn=(a0+an)(C0n+C1n+…+Cnn)=...

(1)由an+1=2Sn+3,得an=2Sn-1+3(n≥2)…(2分)相减得:an+1-an=2(Sn-Sn-1),即an+1=3an,∵当n=1时,a2=2a1+3=9,∴a2a1=3,∴数列{an}是等比数列,∴an=3?3n-1=3n…(5分)(2)∵b1+b2+b3=15,b1+b3=2b2,∴b2=5…(6分)由题意,a13+b1,a23+b2...

an=1+(n-1)×2=2n-1,则1anan+1=1(2n?1)(2n+1)=12(12n?1?12n+1),所以1a1a2+1a2a3+…+1anan+1=12(1-13)+12(13?15)+…+12(12n?1?12n+1)=12(1-13+13?15+…+12n?1?12n+1)=12(1-12n+1)=n2n+1;故答案为:n2n+1.

(1)①由题意,得2a1+4d=178a1+28d=56,解得d=-1…(4分)②由①知a1=212,所以an=232?n,则bn=3n?an=3n?(232?n),…(6分)因为bn+1-bn=2×3n×(10-n)…(8分)所以b11=b10,且当n≤10时,数列{bn}单调递增,当n≥11时,数列{bn}单调递减,故当n...

(1)设{an}的公差为d,则cn+1-cn=(an+12-an+22)-(an2-an+12)=2an+12-(an+1-d)2-(an+1+d)2=-2d2∴数列{cn}是以-2d2为公差的等差数列(4分)(2)∵a1+a3+…+a25=130a2+a4+…+a26=143-13k∴两式相减:13d=13-13k∴d=1-k(6分)∴13a1+13(13?1)2×...

(1)∵点(a8,bn)在函数f(x)=2x的图象上,∴bn=2an,又等差数列{an}的公差为d,∴bn+1bn=2an+12an=2an+1?an=2d,∵点(a8,4b7)在函数f(x)的图象上,∴4b7=2a8=b8,∴b8b7=4=2d,解得d=2.又a1=-2,∴Sn=na1+n(n?1)2d=-2n+n(n?1)2×2=n2-3n....

(1)依题意,a5=b5=b1q5-1=1×34=81,故d=a5-a15-1=81-14=20,所以an=1+20(n-1)=20n-19,令Sn=1×1+21×3+41×32+…+(20n-19)?3n-1,①则3Sn=1×3+21×32+…+(20n-39)?3n-1+(20n-19)?3n,②①-②得,-2Sn=1+20×(3+32+…+3n-1)-(20n-19)?3n=1+20×3(1-3n-1)1...

网站首页 | 网站地图
All rights reserved Powered by www.jcst.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com